Proof of irrationality of square root 2 and square root 3

Although 3 is rational, its square root need not be rational. If \(\sqrt{3}=p/q\), then \(p^2=3q^2\) makes both \(p\) and \(q\) divisible by 3. Tip: test the square root separately.

Assume \(4+\sqrt{3}\) is rational. Since 4 is rational, subtracting it would make \(\sqrt{3}\) rational, contradicting its irrationality. Hence the sum is irrational. Exam tip: adding or subtracting a rational number does not change irrationality.

Substituting \(a=2r\) gives \(a^2=4r^2\). Using this in \(a^2=2b^2\), we get \(4r^2=2b^2\), or \(b^2=2r^2\). Thus, \(b^2\) is even. If the square of an integer is even, the integer itself must be even; hence \(b\) is even. Being negative is not a reason for \(b\) to be even. Exam tip: this result is used to obtain a contradiction in irrationality proofs.

Given \(p=3k\), substitute it into \(p^2=3q^2\): \((3k)^2=3q^2\), so \(9k^2=3q^2\). Dividing by 3 gives \(q^2=3k^2\); hence \(q^2\) is divisible by 3. Since 3 is prime, if the square of an integer is divisible by 3, then the integer itself is divisible by 3. Therefore, \(q\) is divisible by 3. Exam tip: for a prime \(r\), \(r\mid n^2\) implies \(r\mid n\).

In (\sqrt{2}), (2) is the key factor, and in (\sqrt{3}), (3) is the key factor. Identify the difference by the factor.

At the start the fraction is in lowest form so (a) and (b) are assumed coprime. Both even is the final contradiction.

At the start (\frac{p}{q}) is in lowest form. Both divisible by (3) is derived later as a contradiction.

\(\sqrt{2}\) and \(\sqrt{3}\) are irrational because 2 and 3 are not perfect squares. But \(\sqrt{4}=2\) and \(\sqrt{9}=3\) are rational. Exam tip: the square root of a perfect-square integer is an integer.

As \(3\mid p^2\) and 3 is prime, \(3\mid p\). Put \(p=3k\): \(9k^2=3q^2\), so \(q^2=3k^2\) and \(3\mid q\). Thus both share 3, contradicting coprimality. Exam tip: state the prime-divisor rule clearly.

Assume for contradiction that √3 = p/q, where p and q are integers, q ≠ 0, and the fraction is in lowest terms. Squaring gives p² = 3q². This implies that 3 divides p², and therefore 3 divides p; writing p = 3k then gives q² = 3k², so 3 also divides q. Thus p and q have the common factor 3, contradicting the assumption that p/q was in lowest terms. The contradiction does not make √3 an integer, rational, or zero. Instead, it disproves the rational assumption. Therefore √3 is irrational, and option C is the correct final conclusion.

\(\sqrt{2}\) is irrational. If \(5+\sqrt{2}\) were rational, subtracting the rational number 5 would make \(\sqrt{2}\) rational, a contradiction. Hence the sum is irrational. Exam tip: isolate the irrational term by adding or subtracting a rational number.

In the proof of (\sqrt{3}), the chain uses divisibility by (3). It shows (q) is also divisible by (3).

A rational number is written in lowest form as a ratio of coprime integers. The denominator cannot be zero.

When a number is assumed to be rational, it can be represented as p/q with p and q integers, q ≠ 0, and the fraction reduced to lowest terms. Lowest terms means that p and q are coprime, so they have no common divisor greater than 1. This condition is essential because the proof later derives that both are divisible by 3, producing the contradiction. Option A describes the result obtained later, not the initial assumption. Options C and D are impossible because a denominator cannot be zero and p/q would not be a valid reduced representation if both were zero. Therefore option B correctly states the required starting condition.

Writing the fraction in lowest terms makes \(p\) and \(q\) coprime. From \(p^2=3q^2\), 3 divides \(p\), and then \(q\), creating a contradiction. Exam tip: state “lowest terms” explicitly.

A prime factor appears in a square only if it appears in the original number. Therefore (p) is divisible by (3).

The square root of a rational number is not always rational. If \(\sqrt{3}=p/q\) in lowest form, then \(p^2=3q^2\), so 3 divides \(p\) and then \(q\), a contradiction. Exam tip: the square root of a non-perfect-square integer is irrational.

If \(\frac{3+\sqrt{2}}{5}\) were rational, multiplying by 5 and subtracting 3 would make \(\sqrt{2}\) rational, a contradiction. B, C and D equal 2, 1 and 0. Exam tip: use closure of rational numbers to test such expressions.

Assume \(\sqrt{3}=p/q\) in lowest terms. From \(p^2=3q^2\), \(p\) is divisible by 3, and then \(q\) is also divisible by 3. This contradicts coprimality. Exam tip: state “lowest terms” first.

Let \(p=3k\). Substituting in \(p^2=3q^2\) gives \(9k^2=3q^2\), so \(q^2=3k^2\). Hence \(q\) is also divisible by 3, contradicting that \(p\) and \(q\) are coprime. Exam tip: if a prime divides a square, it divides the number itself.